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The positive sign printed in the exponential uses the skew-adjoint convention for the Dirac operator: $D^{\ast }=-D$, so $D_{-}=-D_{+}^{\ast }$. Put $A=D_{+}$ and define the nonnegative self-adjoint operatorsFor $t>0$, the assumed compact elliptic spectral theorem for these elliptic differential operators gives discrete eigenvalues, finite multiplicities and trace-class operators $e^{-tP}$, $e^{-tQ}$.

If $Pu=\lambda u$ with $\lambda >0$, then $Q(Au)=A(Pu)=\lambda Au$ and $Au\ne 0$. Conversely $A^{\ast }$ sends the $\lambda$ eigenspace of $Q$ to that of $P$. On these eigenspaces the inverse of $A$ is ${\lambda }^{-1}{A}^{\ast }$. Thus the positive spectra agree with multiplicity, and absolute convergence of the operator traces permits cancellation term by term:The last expression is the Fredholm index of $A$, since its closed range has orthogonal complement $\ker A^{\ast }$. ConsequentlyThis is the supertrace cancellation used in Question 1. If the Dirac operator is instead defined to be self-adjoint, $D_{-}=(D_{+}{)}^{\ast }$ and the heat exponentials must have a minus sign. With that convention the positive exponentials in the printed formula are generally unbounded and have no finite operator trace.

A Lie algebra representation $V$ is completely reducible when it is a direct sum of irreducible submodules. In finite dimension this is equivalent to saying that every submodule $W\subseteq V$ has an invariant complementary subspace. Indeed, one may split off a minimal nonzero submodule and induct on dimension; conversely, if $V$ is a finite sum of simple submodules, choose a maximal sum $U$ of these with $U\cap W=0$. If $W+U\ne V$, some simple summand $S$ is not contained in $W+U$. Then $S\cap (W+U)=0$ by simplicity, so $U+S$ is a larger sum disjoint from $W$, a contradiction. Thus $V=W\oplus U$.

We prove the required splitting by Haar averaging produces invariant complements. The structural input is the compact real form of a complex semisimple Lie algebra: there is a real semisimple algebra $\mathfrak{k}$ with $L=\mathfrak{k}\oplus i\mathfrak{k}$, whose simply connected integrating Lie group $K$ is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses root vectors normalized so that $[e_{\alpha },e_{-\alpha }]=h_{\alpha }$, with real structure constants compatible with the involution $e_{\alpha }\mapsto -e_{-\alpha }$ and $h_{\alpha }\mapsto -h_{\alpha }$. The real spanis closed under brackets, complexifies to $L$, and has negative-definite Killing form. The compactness criterion for real semisimple Lie algebras then gives a compact simply connected $K$. Thus this route uses root structure and compact integration, not the theorem being proved.

Restrict $\rho :L\to {\mathrm{End}}_{\mathbb{C}}(V)$ to $\mathfrak{k}$. The integration of a Lie-algebra representation gives a representation $R:K\to {\mathrm{GL}}_{\mathbb{C}}(V)$, since $K$ is simply connected. Choose any positive-definite Hermitian inner product $⟨,{⟩}_0$ and average using normalized Haar measure:The integral exists by compactness. It is positive-definite because $R(g)v\ne 0$ for $v\ne 0$, and it is $K$-invariant by translation invariance of Haar measure. Equivalently, every $\rho (a)$ with $a\in \mathfrak{k}$ is skew-Hermitian for this inner product.

Let $W$ be an $L$-submodule. It is invariant under $\mathfrak{k}$, hence under the exponentials generating the connected group $K$. For $v\in W^{\perp }$, $w\in W$ and $g\in K$,Consequently the orthogonal complement $W^{\perp }$ is $K$-invariant, and differentiation makes it $\mathfrak{k}$-invariant. It is a complex vector subspace, so it is also invariant under $i\mathfrak{k}$ and hence under $L$. Therefore $V=W\oplus W^{\perp }$ as $L$-modules.

Induction on $\dim V$ now expresses $V$ as a finite direct sum of irreducible submodules. Every finite-dimensional representation of a finite-dimensional complex semisimple Lie algebra is completely reducible. The compact real form need not be the real form originally used to present $L$; in particular the averaging argument does not require the given presentation to be unitary.

The map $h\mapsto Ph$ is a bounded linear functional on L2 space, since $|Ph|\le \Vert h{\Vert }_2$ by the Cauchy-Schwarz inequality. Its kernel, the mean-zero L2 space, is therefore closed. Every nuisance score function and the parametric score function $\dot{\ell }$ are centered, so both ${\Pi }_{\mathcal{N}}\dot{\ell }$ and $\tilde{\ell }$ are centered. HenceMoreover, the efficient score belongs to the orthogonal complement of the nuisance tangent space. Writing $\dot{\ell }=\tilde{\ell }+{\Pi }_{\mathcal{N}}\dot{\ell }$ givesThis is the efficient-score projection identity; it remains valid when the efficient information is zero.

An orthonormal multiresolution analysis consists of closed subspaces of a Hilbert space $(V_j{)}_{j\in \mathbb{Z}}$ of the Hilbert space $L^2(\mathbb{R})$ such thatThere is a scaling function $\varphi$ whose integer function translations form an orthonormal basis of $V_0$. Consequently ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ form an orthonormal basis of $V_j$; in particular $V_0$ is invariant under integer function translations.

Let $W_j=V_{j+1}\ominus V_j$ be the orthogonal complement of $V_j$ in $V_{j+1}$. Nesting, density and the trivial intersection imply the orthogonal direct sumFor example, telescoping $V_{J+1}=V_L\oplus {⨁}_{j=L}^J{W}_j$ and letting $L\to -\infty$, $J\to \infty$ gives this decomposition; the corresponding orthogonal projections converge strongly to zero and the identity by the intersection and density properties.

To construct the orthonormal wavelet, use $\varphi \in V_1$ to write the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }^2$ and ${\sum }_n|a_n{|}^2=2$. Put $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The orthonormal translates and Fourier periodization criterion proved in part (b), applied to the scaling refinement equation, yieldsThis is the quadrature mirror filter identity. DefineThe Fourier series of $q$ is $\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$, and $\hat{\psi }(2t)=q(t)\hat{\varphi }(t)$. The two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ form a unitary $2\times 2$ matrix almost everywhere: both have squared norm one and their inner product is zero. This wavelet completion of a multiresolution filter proves not just orthogonality but completeness. Splitting a finer-scale coefficient Fourier series into its values at $t$ and $t+\pi$, this invertible two-channel matrix resolves it into a coarse-scale channel and its complementary channel. Thus the integer function translations of $\psi$ form an orthonormal basis of $W_0$.

The dilated functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ therefore form an orthonormal basis of each $W_j$. The orthogonal direct sum proves the conclusion: every orthonormal multiresolution analysis produces an orthonormal wavelet basis $\{{\psi }_{j,k}:j,k\in \mathbb{Z}\}$ of $L^2(\mathbb{R})$.

If $H$ is a finite-dimensional Hilbert space, a choice of orthonormal basis identifies its closed unit ball with a closed bounded ball in a finite-dimensional Euclidean space. The Heine-Borel theorem makes it compact in the norm topology.

Conversely, if $H$ is infinite-dimensional, choose a unit vector $e_1$ and inductively choose $e_{n+1}$ orthogonal to the span of $e_1,\dots ,e_n$. The finite span is a proper closed subspace, so its orthogonal complement contains a nonzero vector that can be normalized. The resulting orthonormal sequence lies in the closed unit ball and satisfiesIt has no norm-convergent subsequence. Sequential compactness is equivalent to compactness in a metric space, hence the closed ball is not compact. This proves the norm-compact unit ball criterion: a Hilbert space has a norm-compact closed unit ball exactly when it is finite-dimensional.

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

Since $\rho (M)=1$, ${\Pi }^{2}f=\rho (f)\rho (M)M=\Pi f$. Also $Lf=0$ exactly when $f=\rho (f)M$, which is exactly the range of $\Pi$. ThusA bounded self-adjoint idempotent is an orthogonal projection: if $u=\Pi u$ and $w=(I-\Pi )w$, then $⟨u,w⟩=⟨u,\Pi w⟩=0$. Here the orthogonal complement is precisely the zero-mass subspace $\rho (f)=0$.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

The intended self-adjoint operator identity is $⟨Lf,g⟩=⟨f,Lg⟩$, with the inner product linear in its first argument. The printed second $Lf$ must be $Lg$. Taken literally, the printed identity forces $L=0$: put $g=0$ and then vary $g$. Its spectral conclusion is then trivial, but it does not define self-adjointness.

For the corrected identity, $⟨Lv,v⟩$ is real. If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality givesThus $L-\lambda I$ is injective and bounded below, so its range is closed: a convergent image sequence has a Cauchy sequence of preimages. Its range is dense because its orthogonal complement is $\ker (L-\bar{\lambda }I)$, also zero by the same lower bound. Hence it is surjective with inverse norm at most $1/|\mathrm{Im}\lambda |$. Therefore

For a bounded self-adjoint operator, $g$ is orthogonal to its range exactly when $⟨Lf,g⟩=⟨f,Lg⟩=0$ for every $f$, or equivalently $Lg=0$. ConsequentlyThis is image-kernel orthogonality for an adjoint. Taking the double orthogonal complement gives the closure; it does not assert that the range itself is closed.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The adjoint operator ${\alpha }^{\ast }:W\to V$ is uniquely characterized byLet $A=\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma :V\to V$. Its quadratic form is nonnegative:If $Av=0$, both norms vanish. In particular $v\in \ker \gamma =\mathrm{im}\beta$, while ${\beta }^{\ast }v=0$ means $v$ is orthogonal to $\mathrm{im}\beta$, by the defining adjoint operator identity. A vector belonging to a vector subspace and its orthogonal complement must be zero. Hence $A$ is injective, and finite dimensionality makes it surjective.

ThusThe same calculation shows it is a positive-definite operator when $V\ne \{0\}$.

Put $q=(1,1,1,1{)}^{\mathsf{T}}$ and arrange the four vectors as the columns of $A=(x-1)I+q{q}^{\mathsf{T}}$. On the one-dimensional vector subspace spanned by $q$, the eigenvalue is $x+3$. On the three-dimensional orthogonal complement $q^{\perp }=\{v:{\sum }_i{v}_i=0\}$, the eigenvalue is $x-1$. Thusand the vectors fail to be a basis exactly when $x=1$ or $x=-3$.

For $x=1$, all four columns equal $q$. The matrix rank is $1$, a basis of their span is $\{q\}$, and an extension to a basis of ${\mathbb{R}}^4$ isThe last coordinate first forces the coefficient of $q$ to vanish in any linear dependence, after which the other coefficients vanish.

For $x=-3$, the matrix rank is $3$ and the span is exactly $q^{\perp }$. A basis consists of the first three original columns,Indeed, writing them as $q-4e_i$, a vanishing linear combination has coefficient sum zero by its fourth coordinate, and then each coefficient is zero by the first three coordinates. Adjoin $q$ to extend this basis to ${\mathbb{R}}^4$, since $q\notin q^{\perp }$.

Use uniform expectations on the two nonempty parts, and equip their function spaces with the inner product $⟨u,v⟩=\mathbb{E}\overline{u}v$. Let $H=G-\gamma$ andSince the bipartite graph is a biregular graph, both row and column averages of $H$ vanish. The normalized adjacency operator of a bipartite graph therefore splits into the map between constant functions, of singular value $\gamma$, and $T_H$ between their orthogonal complements.

There is no singular value larger than $\gamma$. Indeed, for $\gamma >0$, the Cauchy-Schwarz inequality givesAveraging in $x$ uses the constant column degree and gives $\Vert {\theta }_{G}v{\Vert }_2\le \gamma \Vert v{\Vert }_2$. For $\gamma =0$ every operator is zero. Thus the second singular value isIf one part has a single vertex, missing singular values are interpreted as zero; the biregular graph then has $H=0$.

The discrepancy in (i) is exactly $\mathbb{E}H{1}_A{1}_B$. Its supremum over $A,B$ is the cut norm $\delta =\Vert H{\Vert }_{\mathrm{cut}}$. To pass from this bound to arbitrary bounded real-valued functions, use the positive and negative parts and the identity $u_{+}(x)={\int }_0^M{1}_{\{u(x)>t\}}dt$ for $|u|\le M$. Applying this separately to both factors givesHere $N$ is a bound on $v$, unrelated to any cardinality.

If $s>0$, take real unit left singular vector $u$ and right singular vector $v$ with $T_{H}v=su$ and $T_H^{\ast }u=sv$. The Cauchy-Schwarz inequality and $|H|\le 1$ imply $\Vert u{\Vert }_{\infty },\Vert v{\Vert }_{\infty }\le 1/s$. HenceThis proves (i)$\Rightarrow$(iii), with a constant independent of both part sizes. The case $s=0$ satisfies the same conclusion immediately.

Let $V$ be the closure of the simple predictable processes in the Hilbert space $L^2(\mathcal{P},\mu )$. Common refinement of deterministic partitions makes these processes a vector space. To prove density of simple predictable processes for finite measures, suppose $g\in V^{\perp }$. Since $\mu$ is a finite measure, the Cauchy-Schwarz inequality gives $g\in L^1(\mu )$, sodefines a finite signed measure. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a simple predictable process; hence $\nu$ vanishes on these rectangles.

Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this pi-system. Its $\nu$-mass is zero because $1_{\Omega \times [0,n]}$ is simple and converges to $1$ in $L^2(\mu )$ by the dominated convergence theorem. The zero sets of $\nu$ form a lambda-system now that its total mass is zero. The Pi-lambda theorem therefore gives $\nu (E)=0$ for every $E\in \mathcal{P}$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.

Thus $V^{\perp }=\{0\}$ and the orthogonal complement criterion for a closed subspace of a Hilbert space yieldsOmitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.

The given map is the unilateral shift operator. Its adjoint operator is the left shift operator, sinceThe shift is an isometry with trivial kernel and closed range $\{f:f_1=0\}$. Its range has orthogonal complement $\mathrm{span}\{e_1\}$, so every square-summable datum is admissible. Consequently,Indeed $K^{†}K=I$, while $K{K}^{†}f=f-f_1{e}_1$ is the required orthogonal projection. The discarded first component is the residual of the least-squares solution, not an obstruction to the existence of the Moore–Penrose inverse of an operator.

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,